How to find $P(Y<1\mid X=3)$ given joint density of $(X,Y)$?

Question

If $(X,Y)$ has density $f(x,y)=\frac { 1 } { 2 } y ^ { 2 } e ^ { - x } , \text { if } 0 < y < x < \infty\,,$ what is $P(Y<1\mid X=3)$?

First of all if we expand conditional probability formula we will get $P(X=3)$ in denominator and as this is a continuous rv it should be zero?..so it will be undefined.

That's what comes to my mind. But it has a proper answer.

Answer 1

New answer: Give that you said the answer is 1/27, I know my second interpretation is correct. y goes from 0 to x, not from 0 to $\infty$. In this case x =3 given by the statement Probability that y<1 given that x =3

Now we just calculate it as I outlined before: First, normalize the distribution:

$$\int_{0}^{3}\frac{1}{2}y^2 e^{-3} dy = \frac{27}{6e^3}$$ Then just integrate y up to 1 to get the specific probability asked for $$\int_{0}^{1}\frac{1}{2}y^2 e^{-3} dy = \frac{1}{6e^3}$$

Dividing the second integral by the first gives $\frac{1}{27}$

So in general, if it's asking for the probability that it is less than a certain value, you must integrate from the lowest possible value to the value in question. If instead we wanted the probability for y>1 then we would integrate from 1 to 3 instead.

Answer 2

You should look at this probability from the point of view of conditional distributions. Given joint pdf of $X,Y$, you can find conditional pdf of $Y$ given $X$ as $$ f_{Y\mid X}(y\mid x) = \frac{f_{X,Y}(x,y)}{f_X(x)}. $$ Say, for $x=3$, $$ f_{Y\mid X}(y\mid 3) = \frac{f_{X,Y}(3,y)}{f_X(3)}. \tag{1}\label{1} $$ Here the denominator $f_X(3)$ is exactly $$ f_X(3) = \int_{-\infty}^\infty f_{X,Y}(3,y)\,dy = \int_0^3\frac12 y^2 e^{-3}\, dy = \frac{27}{6}e^{-3} $$

And then consider $\mathbb P(Y<1\mid X=3)$ as the probability that a random variable with conditional pdf \eqref{1} takes values less than $1$: $$ \mathbb P(Y<1\mid X=3) = \int_{-\infty}^1 f_{Y\mid X}(y\mid 3)\, dy = \int_0^1 \frac{\frac12 y^2 e^{-3}}{\frac{27}{6}e^{-3}}\, dy = \frac1{27}. $$

Note that there was no need to calculate all constants in conditional pdf. Indeed, it is a function of $y$ only: $$ f_{Y\mid X}(y\mid 3) = \frac{f_{X,Y}(3,y)}{f_X(3)} \propto y^2, \quad 0<y<3. $$ Here $\propto$ means proportional up to a constant. And a normalizing constant is a number $c$ such that $\int_0^3cy^2\,dy=1$. So $$c=\frac{1}{\int_0^3 y^2\,dy}=\frac{1}{9}.$$

Finally the conditional pdf of $Y$ given $X=3$ is $$ f_{Y\mid X}(y\mid 3) = \frac{y^2}{9}\cdot \mathbb 1_{\{0<y<3\}} $$ Finally, $$ \mathbb P(Y<1\mid X=3) = \int_0^1 \frac{y^2}{9} \, dy = \frac{1}{27}. $$

There is a typical thing to consider conditional probabilities or expectations with zero probabilities in the condition part, if we have pair of r.v.'s with continuous joint distribution. This cases are always treated as probabilities for conditional distribution.